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jonny [76]
3 years ago
12

How to find answer for question (a)?​

Mathematics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

i don't have an idea yet please

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\
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\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period}\\
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\end{array}

now.. hmmm let's see, keeping in mind the template above

so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either

and period of 4π, well, that simply means that 

\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\
-----------------------------\\\\

\begin{array}{llll}
f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta  \right)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}
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